Question

If $I_n={\int }_0^{\pi /4}ta{n}^n\theta d\theta$, where n is a positive integer, then $n(I_{n-1}+I_{n+1})$ is equal to

• A. 1
• B. n - 1
• C. $\frac{1}{n-1}$
• D. None of these

Answer 1

Answer: (A)

1

$I_n={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n}dx$

$I_n={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n-2}({\left(\sec x\right)}^2-1)dx$

$I_n={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n-2}.{\left(\sec x\right)}^{2}dx-{\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n-2}dx$

$LetJ={\int }_0^{\frac{\pi }{4}}{\left(\tan x\right)}^{n-2}.{\left(\sec x\right)}^{2}dx$

$Letz=tanx$

$\Rightarrow dz={\left(\sec x\right)}^{2}dx$

$J={\int }_0^1{z}^{n-2}dz$

$J=\frac{1}{n-1}$

$I_n=J-I_{n-2}$

$I_n=\frac{1}{n-1}-I_{n-2}$

$(n-1)I_n+(n-1)I_{n-2}=1$

$n\to n+1$

$\Rightarrow n{I}_{n+1}+n{I}_{n-1}=1$